Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
int x[12345], y[12345];
vector<pair<int, int> > ans;
vector<pair<int, int> > ans2;
void as(int i, int j, vector<pair<int, int> >& ans) {
  ans.push_back(make_pair(i + 1, j + 1));
}
void f(int x[], vector<pair<int, int> >& ans) {
  for (int j = 1 << 30, k = 0; j && k < N; j >>= 1) {
    for (int i = k; i < N; i++)
      if (x[i] & j) {
        if (i != k) {
          swap(x[i], x[k]);
          as(i, k, ans);
          as(k, i, ans);
          as(i, k, ans);
        }
        break;
      }
    if (x[k] & j) {
      for (int i = 0; i < N; i++)
        if (i != k && (x[i] & j)) {
          x[i] ^= x[k];
          as(i, k, ans);
        }
      k++;
    }
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", x + i);
  for (int i = 0; i < N; i++) scanf("%d", y + i);
  f(x, ans);
  f(y, ans2);
  bool ok = true;
  int ref = -1;
  for (int i = 0; ok && i < N; i++)
    if (y[i]) {
      int idx = -1;
      for (int j = ref + 1; j < N; j++)
        if (((y[i] & x[j]) << 1) > y[i] && ((y[i] & x[j]) << 1) > x[j]) idx = j;
      ref = idx;
      for (int j = idx + 1; j < N; j++)
        if (((y[i] & x[j]) << 1) > x[j]) {
          x[idx] ^= x[j];
          as(idx, j, ans);
        }
      if (idx == -1 || x[idx] != y[i])
        ok = false;
      else if (idx != i) {
        swap(x[i], x[idx]);
        as(i, idx, ans);
        as(idx, i, ans);
        as(i, idx, ans);
      }
    }
  if (!ok)
    puts("-1");
  else {
    for (int i = 0; i < N; i++)
      if (x[i] != y[i]) as(i, i, ans);
    printf("%d\n", (int)(ans.size() + ans2.size()));
    for (int i = 0; i < ans.size(); i++)
      printf("%d %d\n", ans[i].first, ans[i].second);
    for (int i = ans2.size() - 1; i >= 0; i--)
      printf("%d %d\n", ans2[i].first, ans2[i].second);
  }
  return 0;
}